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The parity operation in quantum mechanics and quantum field theory is $\hat P|\vec r\rangle=|-\vec r\rangle$, which we can check from the Fourier transform.

Why the spatial inversion operation $P$ in two space dimensions is (x, y) → (−x, y), rather than (x, y) → (−x, -y), based on this definition? Can we interpret the physical action for the 3D action as a reflection about the origin, whereas in 2D is a reflection across the $y$-axis (why not $x$-axis)? Can we read the parity transformation from the tensor $\eta^{\mu\nu}=diag(1,-1,-1)$?

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The parity transformation needs to have a determinant of $-1$. In all dimensions, it is safe to define it as $(x_1, \dots, x_d) \mapsto (-x_1, x_2, \dots, x_d)$ where only one co-ordinate is flipped. Let's call this the good parity transformation because it generalizes well to any $d$.

If $d$ is odd, $(x_1, \dots, x_d) \mapsto (-x_1, \dots, -x_d)$ is also valid and this seems to be what you learned first. I will call this the bad parity transformation because in even $d$ it would just be a rigid rotation with determinant $1$.

The point is that the good and bad parity transformations in odd dimension differ by a rigid rotation. I.e. something with determinant $1$. This also explains why there was no loss of generality in reflecting $x_1$ instead of $x_2$ for example in the good one. So you can quotient $O(d)$ by any of these and get the same subgroup $SO(d)$ as a result.

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  • $\begingroup$ Thanks for the answer! Why both transformations are valid in odd dimensions? Does that mean we have two ways to define $|-\vec r\rangle$? $\endgroup$
    – IGY
    Oct 22, 2023 at 15:51
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    $\begingroup$ There is only one transformation that should be labelled with $\left | -\vec{r} \right >$. But there are infinitely many transformations that have a right to be called parity. The only requirement is determinant -1. $\endgroup$ Oct 22, 2023 at 16:09
  • $\begingroup$ Thank you! So $|-\vec r\rangle$ is the only transformation that flips the eigenvalue, right? We can call this a parity transformation if we, in 3 dimensions, define $(x,y,z)->(-x,-y,-z)$? $\endgroup$
    – IGY
    Oct 22, 2023 at 16:33
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    $\begingroup$ The particular parity transformation you mention is the only one that has $-1$ as its only eigenvalue. The other ones will have eigenvalues of both $1$ and $-1$ where the multiplicity of $-1$ is odd. $\endgroup$ Oct 23, 2023 at 0:39
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  1. Note that there are more reflections than a point reflection, e.g. mirror reflections.

  2. Presumably we are looking for an element $P\in O(n)$ such that $O(n)=SO(n) \sqcup P\circ SO(n)$, i.e. $\det(P) =-1$; preferably $P$ should be an involution.

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  • $\begingroup$ Thanks! Are $(x,y)->(-x,y)$ and $(x,y)->(-x,-y)$ both involution? $\endgroup$
    – IGY
    Oct 22, 2023 at 15:52
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    $\begingroup$ Involution means its own inverse so you can check. $\endgroup$ Oct 22, 2023 at 16:11

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